Problem: Which integral gives the arc length of the graph of $f(x)=\sin(2x)$ over the interval $[0, k]$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^k\sqrt{1+2\cos(2x)}~dx$ (Choice B) B $ \int_0^k\sqrt{1+4\cos^2(2x)}~dx$ (Choice C) C $ \int_0^k\sqrt{1+\sin^2(2x)}~dx$ (Choice D) D $ \int_0^k\sqrt{1+\sin(2x)}~dx$
Explanation: The arc length $L$ of the graph of the function $f$ over the interval $[a, b]$ is $ L = \int_a^b \sqrt{1+\big[f\,^\prime(x)\big]^2} dx$. First, calculate $f\,^\prime(x)$. $\begin{aligned} f(x) &= \sin(2x)\\ \\ f\,^\prime(x) &= 2\cos(2x) \end{aligned}$ Now apply the arc length formula on the interval $[0,k]$ and simplify the integral. $\begin{aligned} L&=\int_0^k\sqrt{1+\big[2\cos(2x)\big]^2}~dx \\\\ &=\int_0^k\sqrt{1+4\cos^2(2x)}~dx \end{aligned}$